# Solve the following

Question:

$\frac{x-1}{3}+4<\frac{x-5}{5}-2$

Solution:

$\frac{x-1}{3}+4<\frac{x-5}{5}-2$

$\Rightarrow \frac{x-1}{3}-\frac{x-5}{5}<-2-4 \quad\left[\right.$ Transposing 4 to the RHS and $\frac{\mathrm{x}-5}{5}$ to the LHS $]$

$\Rightarrow \frac{5(x-1)-3(x-5)}{15}<-6$

$\Rightarrow \frac{5 x-5-3 x+15}{15}<-6$

$\Rightarrow \frac{2 x+10}{15}<-6$

$\Rightarrow 2 x+10<-90$

$\Rightarrow 2 x<-90-10$  [Transposing 10 to the RHS]

$\Rightarrow 2 x<-100$

$\Rightarrow x<-\frac{100}{2}$    [Dividing both the sides by 2 ]

$\Rightarrow x<-50$

Hence, the solution of the given inequation is $(-\infty,-50)$.